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Question 102727: The tens digit of a certain number is 3 more than the units digit. The sum of the squares of the two digits is 117. Find the number.
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let T represent the tens digit. Let U represent the units digit.
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The problem first tells you that the tens digit is 3 more than the units digit. Write this
in equation form as:
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T = U + 3
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Next the problem tells you that if you square each of these digits and add the squares the
result is 117. The tens digit squared is T^2 and the units digit squared is U^2. Add these
two together and set this sum equal to 117. In equation form this is:
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T^2 + U^2 = 117
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But earlier we found that T = U + 3. So in the "squared" equation we can substitute
U + 3 for T to get:
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(U + 3)^2 + U^2 = 117
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Square the U + 3 term and the equation becomes:
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U^2 + 6U + 9 + U^2 = 117
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On the left side the two U^2 terms combine and the equation becomes:
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2U^2 + 6U + 9 = 117
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To get this equation into a form that can be solved, get rid of the 117 on the right side
by subtracting 117 from both sides to change the equation to:
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2U^2 + 6U - 108 = 0
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Simplify this a little by dividing both sides (all terms) by 2 to reduce the equation to:
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U^2 + 3U - 54 = 0
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Notice that this can be factored to:
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(U + 9)(U - 6) = 0
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This equation will be true if either of the factors is zero, because a multiplication
by zero on the left side will cause the entire left side to become zero and therefore equal
to the right side.
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Setting each of the factors equal to zero results in the possible solutions being U = -9
and U = +6. The minus solution makes no sense ... a minus digit in a number????
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So the solution is that the units digit is 6. And the tens digit is 3 more than that, so
the tens digit is 9. The number is 96.
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Check ... square the 9 to get 81 and square the 6 to get 36. The sum of 81 and 36 is 117.
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Everything works out, so our answer is correct.
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Hope this helps you to understand the problem.
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